the title is the question.

There are numerous threads on this topic. If you'd like the details, I suggest you peruse the archives. But in short: one can achieve pretty much any plate count you want, depending upon gradient conditions such as the gradient slope, gradient shape, duration of gradient, etc. This ability to modify the measured plate count at will renders this parameter useful only for comparisons of identical systems. Systems must be very closely matched for this to be viable. Even small differences in system delay volume can result in significant changes in apparent plate count. The generally accepted tool for assessing a gradient analytical system is peak capacity.

Actually, Chris, I'd like to take issue with what you said (although is suspect we both mean the same thing).

You *can* consider the number of theoretical plates. If you express retention in terms of k', the plate number is exactly the same in isocratic and gradient LC:

N = [t0(1+k')/s]^2

{It's hard to play typographical games here, but "s" is intended to be "sigma"; for a Gaussian peak, the baseline width is 4s.}

The catch (which is think is what you were referring to) is that you *can't* measure N conveniently in a gradient system the way you can in an isocratic system. You have to know the k' of the peak at elution (when it exits the column). That, in turn, can't be estimated from a single chromatographic run. For "linear elution strength" gradients (the norm in reversed phase), you need either data from two runs with different gradient steepness, or an estimate of the reversed-phase S value. Even then, the computation is non-trivial.

The peak capacity *can* be computed from a single chromatogram; in that sense, I agree with you that it is a more useful practical measure. On the other hand, the "universality" of the plate count (10,000 plates is 10,000 plates whether isocratic or gradient) makes it (to my mind, anyway) a more attractive measurement.

Your turn! :)

-- Tom

Tom,

You are right, as usual. I was oversimplifying when I said you can't measure efficiency since this is certainly possible if you know the "real" k' of the analyte. The problem is, in my experience, that any attempt at calculating plate count in gradient mode is generally done with the superficial k' of the analyte, treating it as if it were an isocratic separation.

My guess is that the reason this question is posted here so often is that commercially available software packages spit out an efficiency calculation regardless of the validity of the calculation in the case of gradient operation. This provides a strong motivation for a novice to somehow use this information in spite of the fact that the measurement is not a "valid" measure of chromatographic efficiency. My point was only that this information is really not useful unless comparisons are made under closely controlled conditions. I, for one, frequently use the "canned" calculated efficiency value in a gradient mode as a basis for comparing columns on a given instrument with a given gradient. So, I wouldn't say that such comparisons have no value. It's just that one needs to understand the limitations of such a comparison in order to avoid coming to erroneous conclusions when comparing data from different instruments or different methods.

ok guys!!!!!!!!!!! I've got the point!

tks

sorry, do you mind telling me what do you mean by
"That, in turn, can't be estimated from a single chromatographic run"

ką A = (tA-t0)/t0

why it can not be estimated by a single run?

Thanks so much!!!

To answer the previous post (Anonymous the younger?):

k' at elution is twice the average k' during the separation (often symbolized by k*). For a linear gradient in reversed phase, k* is approximately given by:

k* = 1.15*(tG/range)*(F/Vm)*(1/S)

- tG is the gradient time
- range is the difference between initial and final organic solvent fraction (%change/100)
- F is the flow rate
- Vm is the column internal volume
- S is the slope of the linear relationship between isocratic log(k) and fraction organic solvent.

The S value is the catch. Since it's a slope, 2 experiments are required to compute it. You can use 2 isocratic experiments (at two different %B values), but if you're going to go to that much trouble, it's probably easier just to measure the isocratic plate count (the gradient plate count should be about the same). You can also determine it from two gradient runs with appropriate software (the math gets convoluted). I know the DryLab program from the original LC Resources (now part of Rheodyne) does it. I believe the other chromatography modeling packages (from ACD or ChromSword) also have the capability.

If you want to wallow in the math, there's a review of the "Linear Solvent Strength Model of Gradient Elution" written by Lloyd Snyder and John Dolan in vol 38 of the "Advances in Chemistry" series (Marcel Dekker, 1998 -- I think).

Flattery will get you *everywhere*, Chris! :)

Seriously, though, as you point out, both statements are accurate in their own context.

-- tom

yeah, I think maybe younger than you, from the photo on the website :-)

so, less experienced, still learning :-)

Thanks,

jin

By the way, what do you mean, capacity k' AT ELUTION and AVERAGE k' during separation. I think isocratic gradient can also separate mixtures,

And also, what is the meaning of "real" k', and "superficial" k' in Tom's email?

Sorry, maybe "stupid" question, really thanks for answering!

jin

Hey, can I jump in here?

The effeciency equation N=16(t1/w1)^2 works well when the elution is isocratic. If you have a series of peak times (t2, t3, t4..tx) and a series of peak widths (w2, w3, w4..wx) you will find N comes out the same with each computation. If you try the same thing using a gradient, N changes. This is due to changing k' which affects t1.

When I do step gradients you can compare all the peaks on that step to each other in terms of similar N. I teach my people that when you change from Buffer A to Buffer B in a whole step no fade in, you reset the column conditions and get a "new column." I have a buffer that elutes acidic compounds, then a new buffer to elute neutrals, then a third to elute basic compounds. Each time the buffer changes the peak shape resets to sharp and later peaks are broader. This follows the above equation.

I tend to think as constant gradient as constantly changing your initial conditions until the compound hits the one it likes and hops off. Also, this is my favorite chromatography equation ever since LC/GC published years ago how to find the true elution time of broad ghost peaks by plugging in the numbers!

to jin. No, not stupid at all. I think that Lloyd Snyder (the "father" of linear solvent strength gradient theory) had to explain it to me a dozen times or so before comprehension slowly seeped through my skull.

Yes, many mixtures can be separated isocratically. In many other cases, however, there is too wide a range of retention to fit into a single isocratic run (by the time you get the mobile phase weak enough to have reasonable retention for your first peak, the later peaks are retained far to long). That's what gradients are used for.

k' is the ratio of the number of analyte molecules in the stationary phase to the number of analyte molecules in the mobile phase. It is fundamentally related to the distribution coefficient of the analyte between the two phases (in fact, it's the distribution coeffient multiplied by the phase ratio). As long as the mobile phase composition stays the same (isocratic), then the distribution coefficient stays the same, and there is no problem measuring k'. If the mobile phase composition changes, then the distribution coefficient changes, and k' changes. If you then try to use the isocratic formula, what you will get is an "apparent" or "superficial" k' which doesn't relate to the distribution coefficient.

Conceptually, the simplest gradients are "step" gradients (e.g., run at 10% organic for 10 minutes, and then jump to 25% organic). As the previous post points out, each of those segments represents a "new" set of conditions, and you can pretty much calculate k' values (and plate counts, for that matter) for each segment separately (in real life, instrument mixing volume issues complicate things a bit). The catch is that you have to have a reasonable "break point" in your chromatography where nothing of interest is eluting from the column.

As a matter of practice, most gradients are linear ramps of strong solvent concentration, and this is what most people refer to when they use the word "gradient". In this situation, the k' of a peak gradually decreases as it moves down the column during the course of a gradient The peak will move slowly (i.e., have a large k') when it is still near the column inlet, and will be moving more quickly (i.e., have a low k') as it approaches the column outlet). We can define the "average" k' for a peak as the k' value of that peak in the mobile phase composition that occurs when it is at the midpoint of the column. That value is called k*.

The k' of a peak continues to decrease as it moves down the column past the midpoint, because the mobile phase strength continues to increase. The k' of the peak when it is just at the column exit ("k at elution") will determine the peak width.

Sorry to be so long-winded!

-- Tom

To Tom. very clear explaination, thanks so much, I will make it DEEPLY seep through my skull, thanks,

jin

hi, Tom,

In the reply in Wednesday, May 5, 2004 - 1:30pm. There is once sentence:

The S value is the catch. Since it's a slope, 2 experiments are required to compute it. You can use 2 isocratic experiments (at two different %B values), but if you're going to go to that much trouble, it's probably easier just to measure the isocratic plate count (the gradient plate count should be about the same).

You said: THE GRADIENT PLATE COUNT SHOULD BE ABOUT THE SAME.

However, I read another conversation this morning, I attached in the following:

Plate counts of 500,000 per metre ?

By Anonymous on Thursday, May 29, 2003 - 05:37 pm:
Our experience of plate counts is that 5 micron ODS column gives up to 100,000 plates per metre when tested using isocratic conditions.
Recently an associate showed me an example of a plate count of 100,000 for a 15cm column - that is 600,000 plates per metre ( used millenium software). Hand calculation gave same result.
This was a gradient separation - 10% ACN to 90%ACN in 20 mins - is this why the plate count is so unbelievably high ?.


--------------------------------------------------------------------------------
By Chris Pohl on Thursday, May 29, 2003 - 06:35 pm:
Yes. See previous postings on this topic.

So, should the plate counts of isocratic and gradient the same? Or what you really mean is PEAK CAPACITY

Thanks so much for explaination!

jin

The plate count, N, for a given compound on a given column should be about the same whether the separation is carried out under isocratic or gradient conditions (this is an oversimplification, in that it ignores second-order effects like gradient band compression and extra-column volume but it's good enough for practical purposes). The starting point of the thread was the observation that you can't *measure* the plate count the same way in isocratic and gradient systems (in fact, you can't measure the plate count at all from a single gradient run; you need data from two runs).

The peak capacity *can* be measured conveniently from a single gradient run, and in that sense is more useful. It doesn't relate as well to isocratic separations, so in that sense is less useful. "Ya pays yer money and ya takes yer choice!

Despite the hard-core theory that has been discussed I think there are two issues that have not yet been covered, with respect to determining plate counts in gradient mode.

First, the basic defenition of theoretical plates involves the ratio of retention time to peak width. And I think it was defined this way to "normalize" for the fact that peaks get broader as the retention time increases as there is more time for bandbroadeing to occur. But this relationship does not really apply in gradient chromatography. With gradient separations the peak widths are generally fairly constant (because the each component remains focussed on the head of the column until it's turn to migrate).

The second issue is the dwell volume effect. I think you all know what I mean: the time it takes between when the gradient changes and when the column actually sees this change. Systems with larger dwell volumes will result in peaks with larger retention times and, therefore, a larger result for plate count will be obtained. So as a result of this issue the calcuated plate count will vary depending on what system is used.

In my view it is virtually impossible, in gradient mode, to calculate a plate count value that can be used as an indicator of the peformance of the column and use that value to compare to other columns etc. But I think you can always used plate count as a way of monitoring the peak width as compared to what it's been historically (i.e. as a system suitability criteria). But it must be thought of with respect to a given peak on a given column with a given set of conditions and on a given system (different systems are OK if the dwell volumes are similar).

That's the point that a lot of people don't get: the basic definition of theoretical plates does *not* involve a ratio of peak width to retention time; that just happens to be the way in which the measurement is done for an isocratic system.

The basic defition is the number of discrete equilibrations between stationary and mobile phases that would be required to generate the equivalent peak shape. That definition is equally applicable to both isocratic and gradient separations. As it happens, it's easy to measure from a single isocratic run, but requires two gradient runs and some convoluted math.

You're quite right on the effect that dwell volume has if you use the isocratic plate count measurement formula on a gradient system. That just reinforces the point of this whole thread: ya can't measure plates that way!

Tom,

In your post of May 5 you give a formual for working out the number of theoretical plates. You then go on to say that the k' value in this expression is the k' of the peak at elution. On May 6 you said that k gradually decreases as the peak moves down column. On May 5 in your second post you said that k' at elution is twice the average value of k*. I don't understand this at all.

No wonder. Should have been *half*, not twice!

That's what I get for posting late at night!